# 12.09 Review: Completing the square

Lesson

Completing the square is the process of rewriting a quadratic expression from the expanded form $x^2+2bx+c$x2+2bx+c into the square form $\left(x+b\right)^2+c-b^2$(x+b)2+cb2. The name comes from how the method was first performed by ancient Greek mathematicians who were actually looking for numbers to complete a physical square.

There are two ways to think about the completing the square method: visually (using a square), and algebraically. We will see that we can use this method to find the solutions to the quadratic equation and to more easily graph the function.

### Perfect square trinomials

The basis for completing the square relies on the factoring technique for perfect square trinomials.

Remember!

A perfect square trinomial is of the form $a^2+2ab+b^2$a2+2ab+b2 and factors to give $\left(a+b\right)^2$(a+b)2

For the most part, we will be using that:

$x^2+2bx+b^2=\left(x+b\right)^2$x2+2bx+b2=(x+b)2

or

$x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2$x2+bx+(b2)2=(x+b2)2

Let's practice factoring perfect square trinomials before we start completing the square.

#### Practice questions

##### Question 1

Complete the distribution of the perfect square: $\left(x-3\right)^2$(x3)2

1. $\left(x-3\right)^2=x^2-\editable{}x+\editable{}$(x3)2=x2x+

##### Question 2

Factor $x^2-12x+36$x212x+36.

##### Question 3

Factor $x^2-16x+64$x216x+64.

### Visually completing the square

$x^2+4x-5$x2+4x5

$x^2+2x-3$x2+2x3

$x^2-8x-20$x28x20

Let's imagine we want to factor them using the complete the square method. Now, none of these are perfect square trinomials, but we are going to add or subtract a constant to make it a perfect square trinomial.

The steps involved are:

• Move the constant term to the end
• Create a box and complete the square
• Keep the expression balanced
• Write the factoring

As this is a visual method, its best to learn by watching the process in action, so here are two examples.

#### Worked examples

##### Question 4

We wish to factor $x^2+4x-5$x2+4x5 by first completing the square.

Completing the square on $x^2+4x-5$x2+4x5:

Then can also show this algebraically.

 $x^2+4x-5$x2+4x−5 $=$= $x^2+4x+4-4-5$x2+4x+4−4−5 $=$= $\left(x+2\right)^2-4-5$(x+2)2−4−5 $=$= $\left(x+2\right)^2-9$(x+2)2−9

##### Question 5

We wish to factor $x^2-8x-20=0$x28x20=0 by first completing the square.

Completing the square on $x^2-8x-20$x28x20:

We can also factor this algebraically.

 $x^2-8x-20$x2−8x−20 $=$= $x^2-8x+16-16-20$x2−8x+16−16−20 $=$= $\left(x-4\right)^2-16-20$(x−4)2−16−20 $=$= $\left(x-4\right)^2-36$(x−4)2−36

This interactive will help you to visualize the process. Watch this video for an explanation .

### Completing the square algebraically

Let's imagine we want to factor the same quadratics as before, but this time using the algebraic version of the complete the square method.

#### Exploration

Suppose we wanted to complete the square for $x^2-8x+13$x28x+13. If we complete the square, we want to get it to the form $\left(x-h\right)^2+k$(xh)2+k

The first question is, what will the value of $h$h, the constant in the parentheses, be?

$x^2-8x+13=\left(x-h\right)^2+k$x28x+13=(xh)2+k

If we go back to the idea that $x^2+2bx+b^2=\left(x+b\right)^2$x2+2bx+b2=(x+b)2, we will notice that the constant in the parentheses is half of the coefficient of the linear term. The constant in the parentheses is therefore $\frac{-8}{2}$82, which is $-4$4.

$x^2-8x+13=\left(x-4\right)^2+k$x28x+13=(x4)2+k

Now we need to determine what the value of $k$k will be. To do that, let's consider what $\left(x-4\right)^2$(x4)2 distributes out to.

$\left(x-4\right)^2=x^2-8x+16$(x4)2=x28x+16

Notice that we already had the $x^2$x2 and the $-8x$8x terms, but we had $+13$+13, not $+16$+16 as our coefficient. To account for this, we will add and subtract $16$16, to get what we need, while keeping everything balanced.

$x^2-8x+13=x^2-8x+16-16+13$x28x+13=x28x+1616+13

Now, by collecting like terms, we are all done!

 $x^2-8x+13$x2−8x+13 $=$= $x^2-8x+16-16+13$x2−8x+16−16+13 Adding and subtracting $\left(\frac{-8}{2}\right)^2=16$(−82​)2=16 $=$= $\left(x^2-8x+16\right)-16+13$(x2−8x+16)−16+13 Consider that we have now created a Perfect Square Trinomials $=$= $\left(x-4\right)^2-16+13$(x−4)2−16+13 Factor the first three terms $=$= $\left(x-4\right)^2-3$(x−4)2−3 Combine Like Terms

The steps involved are:

1. Move the constant term to the end, leaving space for two numbers, $x^2+bx+c$x2+bx+c
2. Add and subtract half the coefficient of $x$x squared, $x^2+bx+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c$x2+bx+(b2)2(b2)2+c
3. Factor the first three terms
4. Combine Like Terms

#### Worked Examples

##### Question 1

What should be added to $x^2-x$x2x to make it a perfect square?

##### Question 2

Using the method of completing the square, rewrite $x^2+3x+6$x2+3x+6 in the form $\left(x+b\right)^2+c$(x+b)2+c.

##### Question 3

Using the method of completing the square, rewrite $x^2-7x+14$x27x+14 in the form $\left(x+b\right)^2+c$(x+b)2+c.